Let a,n be positive integers, and let p be a prime. Prove if
(2^p) + (3^p) = (a^n), that n=1
Well, it is true if p is an odd prime and if by "last digit" he meant the units' digit:
$\displaystyle 2^2=-1\!\!\pmod 5\Longrightarrow 2^{p=2n+1}=\left\{\begin{array}{rl}-2\!\!\!\pmod 5&\,\mbox{ if }n\mbox{ is odd}\\2\!\!\pmod 5&\,\mbox{ if }n\mbox{ is even}\end{array}\right.$ , and the same exactly
happens with $\displaystyle 3$ instead of $\displaystyle 2$ , so in any case (with odd prime) the sum $\displaystyle 2^p+3^p=0\!\!\pmod 5$.
Tonio
2^5+3^5=275
275=5*5*11$\displaystyle \neq{a}^{n}$, n>1
i proved for any prime p>5, n=1
Basically the proof runs like this: LHSmod(25)=0 iff the its last two digits$\displaystyle \in${00,25,50,75}. Define B={00,25,50,75}. Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Consider f(k)=LT(2^k+3^k) for k=1,2,3,... Then we find f(k)$\displaystyle \in$B iff k=20m+5 or k=20m+15, for m=0,1,2,3,... Hence k cannot be a prime if LHSmod(25)=0. Hence k>5 cannot be a prime if n>2 on the RHS.
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EDIT: RED LETTERS
Now for the claim f(k)$\displaystyle \in$B iff k=20m+5 or k=20m+15, for m=0,1,2,3,... made above. See the attachment. All should be clear
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Edit: the smallest loop for LT is of length 20.
EDIT:
In the proof, (2^p) + (3^p) = (a^n) $\displaystyle \Rightarrow$ n=1 is true if {p}={odd numbers}-{20m+5}-{20m+15} because for these p LHSmod(25)$\displaystyle \neq$0. So the set of p's for which n=1 is true is strictly larger than the prime set.
Actually, I suspect that n=1 is true for ANY positive integer p! However a proof for that might be much more involved and I shall not attempt to prove it
Of course this proves the question!! and more than the question!
Let me try to clarify my reasoning a little for you:
1) $\displaystyle {2}^{p}+{3}^{p}=0mod(5)$ for p=odd. We've shown this, right?
2) Now RHS$\displaystyle ={a}^{n}. $If RHS=LHS, then RHS=0mod(5). If n>2, we must also have RHS=0mod(25). So LHS=0mod(25). So if we can show LHS$\displaystyle \neq$0mod(25) for prime p's, then we are done (this is also what abhishekkgp claimed up there).
3) But x=0mod(25) iff last two digits of x are 00, 25, 50 or 75. But LHS$\displaystyle ={2}^{p}+{3}^{p}$ has those last two digits iff p=20m+5 or p=20m+15, which are NOT primes! Hence prime for prime p LHS$\displaystyle \neq0(mod25). $and we're done!
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Why the table is enough? Because there's only FINITE p's we need to check (because we only care about last two digits!)
EDIT: Red letters
Define LT(x)=last two digits of x (so LT(100)=00, LT(234)=34, ect.). Then LT(2^5+3^5)=75. Define $\displaystyle {k}^{*}$ to be the smallest integer>5 such that 1) LT(2^$\displaystyle {k}^{*} $+3^$\displaystyle {k}^{*} $)=00 or 25 or 50 or 75; 2) LT(2^$\displaystyle {k}^{*} $)=LT(2^5); 3) LT(3^$\displaystyle {k}^{*} $)=LT(3^5). Such a $\displaystyle {k}^{*} $ is found by inspecting the first 20 terms of the table to be $\displaystyle {k}^{*} =25$. And LT(2^$\displaystyle {k}^{*} $+3^$\displaystyle {k}^{*} $)=75. No problem with this?
Now because LT function takes only the last two digits, it is true that LT(2^(5+i))=LT$\displaystyle ({2}^{5}{2}^{i})$=LT(LT(2^5)LT(2^i))=LT(LT(2^$\displaystyle {k}^{*}$)LT(2^i))=LT($\displaystyle {2}^{{k}^{*}}{2}^{i}$)=LT($\displaystyle 2^{{k}^{*}+i}$), for any i>0. Similarly LT(3^(5+i))=LT($\displaystyle 3^{{k}^{*}+i}$), too. So LT(2^(5+i)+3^(5+i))=LT(LT(2^(5+i))+LT(3^(5+i)))=LT (LT(2^($\displaystyle {k}^{*}+i$))+LT(3^($\displaystyle {k}^{*}+i$)))=LT(2^($\displaystyle {k}^{*}$+i)+3^($\displaystyle {k}^{*}$+i))
From this we can see f(k)=LT(2^k+3^k) takes only FINITELY MANY values for k=5,6,7,...
and the values it can take are exactly equal to this set:{f(5),f(6),f(7),...,f($\displaystyle {k}^{*}-1$)}, and any f(k)$\displaystyle \in${00,25,50,75} iff there exists 5$\displaystyle \leq$k'$\displaystyle \leq$$\displaystyle {k}^{*}-1$ such that f(k')=f(k).
So what is this k' if f(k')$\displaystyle \in${00,25,50,75}? Inspect the table we find k'=5 or k'=15.
So now you see where k=5+20m and k=15+20m come from.
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If this doesn't convince you. I happily quit
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EDIT: red letters.